Question

The vertical component of earth's magnetic field at a place is $\sqrt{3}$ times the horizontal component the value of angle of dip at this place is?

• A. ${30}^o$
• B. ${45}^o$
• C. ${60}^o$
• D. ${90}^o$

Answer 1

Answer: (C)

${60}^o$

As, $B_V=\sqrt{3}{B}_H$

Also, $\tan \delta =\frac{B_V}{B_H}=\frac{\sqrt{3}{B}_H}{B_H}=\sqrt{3}$

or $\delta ={\tan }^{-1}\left(\sqrt{3}\right)={60}^o$

$\therefore$ Angle of dip, $\delta ={60}^o$.